Solveeit Logo
Login

Question

Question: \({(C{H_3})_3}C - O - C{H_3}\) on reaction with \(HI\) gives \({(C{H_3})_3}C - I\) and \(C{H_3}OH\) ...

(CH3)3C−O−CH3{(C{H_3})_3}C - O - C{H_3} on reaction with HIHI gives (CH3)3C−I{(C{H_3})_3}C - I and CH3OHC{H_3}OH as the main product and not (CH3)3C−OH{(C{H_3})_3}C - OH and CH3−IC{H_3} - I.

Explanation

Solution

The given is an example of reaction of ether with HIHI . One degree alcohol and alkyl iodide are formed. Here,the alkyl group is tertiary , the tertiary halide will form because the departure of the leaving group will create a tertiary carbocation and reaction will follow the SN1S{N_1} mechanism .

Complete step-by-step answer:
Any chemical reaction follows a few basic steps. These are :
Protonation: Addition of H+{H^ + }.
(CH3)3C−O−CH3+HI→H+(CH3)3COCH3+{(C{H_3})_3}C - O - C{H_3} + HI\xrightarrow{{{H^ + }}}{(C{H_3})_3}COC{H_3}^ +
Formation of the carbocation : tertiary carbocation will be formed because it is most stable.
(CH3)3COCH3+→CH3OH+(CH3)3C+{(C{H_3})_3}COC{H_3}^ + \xrightarrow{{}}C{H_3}OH + {(C{H_3})_3}{C^ + }
Nucleophilic attack : I−{I^ - } is a good nucleophile and a tertiary carbocation is a great electrophile. Both will add with each other and form tertiary iodide.
(CH3)3C++I−→(CH3)3CI{(C{H_3})_3}{C^ + } + {I^ - } \to {(C{H_3})_3}CI

These three are the basic steps in any chemical reaction. When ether and HIHI react with each other , primary alkyl iodide and primary alcohol are formed . The reaction proceeds through the SN2S{N_2} mechanism.

I−{I^ - } is a good nucleophile . It displaces −OH - OH molecule by SN2S{N_2} mechanism. If the primary and secondary alkyl group are present , the lower alkyl group forms the halide followed by SN2S{N_2}. This is the general case of reaction. But when the alkyl halide is tertiary , the reaction proceeds through the SN1S{N_1} mechanism and hence the product is different from expected.

Note: The above reaction has proceeded through nucleophilic substitution of first order. In theSN1S{N_1} mechanism , the rate of reaction is fastest for tertiary carbocation. Hence , In step 2, the formation of tertiary carbocation is favourable and forms the basis of the final product.